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marusya05 [52]
3 years ago
9

FOR LondonCreamCakes no sure if its right but im trying to help.. lol dont mind all the tabs

Chemistry
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

It kinda helps but not  really

Thanks for trying anyway doe!

Explanation:

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A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

3 0
3 years ago
80cm^3 of oxygen gas diffused through a porous hole in 50 seconds how long wiil it take 120cm^3 of nitrogen (iv) oxide to diffus
inysia [295]

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Explanation:Please see attachment for explanation

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3 years ago
How do the particles that are dispersed in a colloid differ from the particles suspended in a suspension?
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4 0
3 years ago
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